The longest imaginable

This may not be the first time for you to see this solution.

First, we note of this integral

$\displaystyle \int_0^{\infty} e^{-ax} \mathrm{d}x = \frac{1}{a}$

We can actually use this to our advantage, to transform the integrand in question, to an integral wrt to a dummy variable $t$ , that is

$\dfrac{n}{x_1 + x_2 + x_3 + \dots + x_n} = \displaystyle\int_0^{\infty} e^{ - \frac{x_1+x_2+\dots+x_n}{n}t} \mathrm{d}t$

Plugging this into the equation, we will get

$\displaystyle \int_0^1 \dots \int_0^1 \mathrm{d}x_1\mathrm{d}x_2 \mathrm{d}x_3\dots \mathrm{d}x_n \int_0^{\infty} e^{ -\frac{x_1+x_2+\dots+x_n}{n}t} \mathrm{d}t$

And changing the order of integration, we'll get this

$\lim\limits_{n\to \infty} \displaystyle \int_0^{\infty} \bigg( \int_0^1 e^{-\frac{x}{n} t} \mathrm{d}x \bigg)^n \mathrm{d}t$

$\lim\limits_{n \to \infty} \displaystyle \int_0^{\infty} \bigg( \frac{ 1 - e^{ -t/n}}{t/n} \bigg)^n \mathrm{d}t$

$\displaystyle \int_0^{\infty} \lim\limits_{n \to \infty} \bigg( \frac{ 1 - e^{ -t/n}}{t/n} \bigg)^n \mathrm{d}t$

$\displaystyle \int_0^{\infty} e^{-t/2} \lim\limits_{n \to \infty} \bigg( \frac{ e^{ t/2n} + e^{-t/2n}}{t/n} \bigg)^n \mathrm{d}t$

$\displaystyle \int_0^{\infty} e^{-t/2} \lim\limits_{n \to \infty} \bigg( \frac{2 \sinh (t/2n)}{t/n} \bigg)^n \mathrm{d}t$

$\displaystyle \int_0^{\infty} e^{-t/2} \lim\limits_{n \to \infty} \bigg( \frac{ \sinh(t/2n)}{t/2n} \bigg)^n \mathrm{d}t$

Now, note that $\lim_{k\to 0} \frac{\sinh kx}{kx} = 1$

This simplifies our integral to

$\displaystyle \int_0^{\infty} e^{-t/2} \mathrm{d}t = 2$ . (See edit below)

Edit: to justify for the said limit, we will isolate it first and solve it by means of L'Hopital's rule. First we let $u = \frac{t}{2}$ .

$A = \lim\limits_{n \to \infty} \bigg[ \frac{ \sinh(u/n)}{ u/n} \bigg]^n$

Since A is of the indeterminate form $1^{\infty}$ , we need to transform it to a form where L'Hopital's rule can be used.

$\ln A = \lim\limits_{n \to \infty} n \ln \bigg[ \frac{ \sinh(u/n)}{u/n} \bigg]$

This is now of the form $0 \cdot \infty$ . A little more modification will lead us to

$\ln A = \lim\limits_{n \to \infty} \dfrac{ \ln \bigg( \frac{\sinh(u/n)}{u/n} \bigg) }{ \frac{1}{n}}$

This can now be evaluated via L'hopital's rule. Differentiating both the numerator and the denominator twice wrt $n$ , we will get

$\ln A = \lim\limits_{n \to \infty} \dfrac{ n^3 [ u^2 \text{csch}^2 (u/n) - 2un \coth(u/n) +n^2] }{n^4}$

This will evaluate to zero.

Going back now, we can say that $A = e^0 = 1$ .

Let $\{X_i\}_{i}$ be a sequence of i.i.d. random variables each following a uniform distribution in the interval $[0,1]$ , with expectation $\mu=1/2$ . Then by the Strong Law of Large Numbers , we have almost surely $\lim_{n \to \infty} \frac{n}{X_1+X_2+\ldots + X_n} \to \frac{1}{\mu}.$ Then, assuming that the limit and expectation can be exchanged (usually proved by showing that the random sequence is Uniformly Integrable, which I skip), we have $\lim_{n} \mathbb{E}(\frac{n}{X_1+X_2+\ldots + X_n} ) = \mathbb{E}(\lim \frac{n}{X_1+X_2+\ldots + X_n} )=\frac{1}{\mu}=2.$ The proof follows after noting that the LHS is simply the integral in question. $\blacksquare$